I figured out everything but i cant figure out this last one below. Any ideas? Thanks in advance.
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Originally Posted by Evan.Kimia I figured out everything but i cant figure out this last one below. Any ideas? Thanks in advance. $\displaystyle P(t) = 120e^{.804718956218t}$ $\displaystyle P'(t) = 120*.804718956218e^{.804718956218t}$ $\displaystyle P'(5) = 120*.804718956218e^{.804718956218* 5}$ $\displaystyle P'(5)=5397.58 \frac{bacteria}{hour}$
Ah, thanks. I did that before and must of typed it wrong on the calc
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